Problem with Geometric Progression Question: An infinite geometric progression has a finite sum. Given that the first term is 18 and that the sum of the first 3 terms is 38. Calculate the values of (i)the common ratio, (ii) the sum to infinity.
What I'm doing:
Sum $s= \cfrac{a(r^n -1)}{r-1}$
$38  = 18\cfrac{r^3 -1}{r-1}$
$38r - 38 = 18r^3 - 18$
$18r^3 - 38r + 20 = 0$
???
 A: i)
The factor theorem tells us that $r-1$ is a factor. So you can use long division to get the quadratic factor: $$
\require{enclose}
\begin{array}{r}
   18r^2+18r-20  \\[-3pt]
r-1 \enclose{longdiv}{18r^3+0r^2-38r+20} \\[-3pt]
\underline{18r^3-18r^2}\phantom{2} \\[-3pt]
                    -20r+20  \\[-3pt]
         \underline{0}
\end{array}
$$
$$18r^3 - 38r + 20 = 0$$
$$\implies (r-1)(18r^2+18r-20)=0$$
$$\implies (r-1)\left(r-\frac23 \right)\left(r+\frac53 \right)=0$$
$$\implies r=1,-\frac53,\frac23$$ but since $-1 \lt r \lt 1 \implies r=\frac23$
ii)
$S_{\infty}=\cfrac{a}{1-r}=\cfrac{18}{1-\frac23}=54$
A: You have not used the fact that $r-1$ is a factor of your polynomial, which will get you a quadratic.  For three terms it is easier to write $18(1+r+r^2)=38$ and you start with the quadratic.  To have a finite sum you need $|r|\lt 1$, which gives you which root to use.
A: $\textbf{hint}$
The first $3$ terms are
$$
a+ar+ar^2 = 38
$$
where $a=18$ now we have to sum to infinity we utilize a well known fact
$$
\sum_{n=0}^{\infty}ar^n = \frac{a}{1-r}
$$ 
where $|r|<1$ which is important for the first calculation for finding $r$.
